Two uniform 79.8-g marbles 1.83 cm in diameter are stacked as shown in the figur

Question

Two uniform 79.8-g marbles 1.83 cm in diameter are stacked as shown in the figure (Figure 1) in a container that is 2.82 cm wide.

Part A

Find the force that the container exerts on the marble at the point of contact A.

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Part B

Find the force that the container exerts on the marble at the point of contact B.

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Part C

Find the force that the container exerts on the marble at the point of contact C.

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Part D

What force does each marble exert on the other?

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Figure 1 of 1

Two uniform 79.8-g marbles 1.83 cm in diameter are stacked as shown in the figure (Figure 1) in a container that is 2.82 cm wide.

Part A

Find the force that the container exerts on the marble at the point of contact A.

NA =   N  

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Part B

Find the force that the container exerts on the marble at the point of contact B.

NB =   N  

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Part C

Find the force that the container exerts on the marble at the point of contact C.

NC =   N  

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Part D

What force does each marble exert on the other?

N =   N  

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Figure 1 of 1

A B

Explanation / Answer

b)
Taking the two marbles as the system, following forces act on them :
(1) their total weight, 2*0.0798*9.8 N downwards
(2) horizontal reactive forces at A and C
(3) Reactive force that the container exerts on the marble at the point of contact B, upwards = R
=> R
= 2*0.0798*9.8 N
= 1.564 N

c) Consider the upper marble. Three forces act on it.
(1) its weight, 0.782 N downwards
(2) Horizontal reactive force, F' at C
(3) Reactive force, F, from the lower marble at the point of contact in a direction passing through the center of the upper marble. Horizontal distance between C and center of the upper marble = (1/2) (2.82 - 1.83) = 0.495 cm

Consider a right triangle having line joining the centers of the marbles as hypotenuse and horizontal line through the center of lower marble and a vertical line from the center of the upper marble as the other sides.
hypotenuse = 1.83 cm
horizontal line = 2.82 - 1.83 = 0.99 cm
vertical line = [(1.83)^2 - (0.99)^2] = 2.08
Vertical component of F
= F * (2.08/1.83) = 1.1369F
This is equal to the weight of the upper marble
=> 1.1369F = 0.782
=> F
= 0.688 N
d) Reaction at C,
F'
= F * (0.99) /(1.83)
= 0.688 * (0.99) /(1.83)
= 0.372 N

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